22

我想通过时间戳执行移动平均。我有两列:温度和时间戳(时间-日期),我想根据每 15 分钟的连续温度观测值执行移动平均值。换句话说,选择数据以基于 15 分钟的时间间隔执行平均。此外,对于不同的时间序列,可以有不同数量的观察。我的意思是所有窗口大小都相等(15 分钟),但每个窗口中可能有不同数量的观察值。例如:对于第一个窗口,我们必须计算 n 次观察的平均值,而对于第二个窗口,我们必须计算 n+5 次观察的平均值。

数据样本:

ID时间戳温度
1 2007-09-14 22:56:12 5.39
2 2007-09-14 22:58:12 5.34
3 2007-09-14 23:00:12 5.16
4 2007-09-14 23:02:12 5.54
5 2007-09-14 23:04:12 5.30
6 2007-09-14 23:06:12 5.20
7 2007-09-14 23:10:12 5.39
8 2007-09-14 23:12:12 5.34
9 2007-09-14 23:20:12 5.16
10 2007-09-14 23:24:12 5.54
11 2007-09-14 23:30:12 5.30
12 2007-09-14 23:33:12 5.20
13 2007-09-14 23:40:12 5.39
14 2007-09-14 23:42:12 5.34
15 2007-09-14 23:44:12 5.16
16 2007-09-14 23:50:12 5.54
17 2007-09-14 23:52:12 5.30
18 2007-09-14 23:57:12 5.20

主要挑战:

我如何学习代码以每 15 分钟进行一次区分,而由于采样频率不同而没有准确的 15 分钟时间间隔。

4

3 回答 3

11

您可以自己加入您的桌子:

select l1.id, avg( l2.Temperature )
from l l1
inner join l l2 
   on l2.id <= l1.id and
      l2.Timestamps + interval '15 minutes' > l1.Timestamps
group by l1.id
order by id
;

结果

| ID |            AVG |
-----------------------
|  1 |           5.39 |
|  2 |          5.365 |
|  3 | 5.296666666667 |
|  4 |         5.3575 |
|  5 |          5.346 |
|  6 | 5.321666666667 |
|  7 | 5.331428571429 |

注意:只做“努力工作”。您应该将结果与原始表连接或附加新列以进行查询。我不知道您需要的最终查询。调整此解决方案或寻求更多帮助。

于 2012-12-11T11:00:25.017 回答
9

假设您想在每 15 分钟间隔后重新启动滚动平均值:

select id, 
       temp,
       avg(temp) over (partition by group_nr order by time_read) as rolling_avg
from (       
  select id, 
         temp,
         time_read, 
         interval_group,
         id - row_number() over (partition by interval_group order by time_read) as group_nr
  from (
    select id, 
           time_read, 
           'epoch'::timestamp + '900 seconds'::interval * (extract(epoch from time_read)::int4 / 900) as interval_group,
           temp
    from readings
  ) t1
) t2
order by time_read;

它基于Depesz 的按“时间范围”分组的解决方案:

这是一个 SQLFiddle 示例:http ://sqlfiddle.com/#!1/0f3f0/2

于 2012-12-11T12:32:45.003 回答
4

这是一种利用该工具将聚合函数用作窗口函数的方法。聚合函数将最后 15 分钟的观察值与当前运行总数一起保存在一个数组中。状态转换函数将落后于 15 分钟窗口的元素从数组中移出,并推送最新的观察结果。最后一个函数简单地计算阵列中的平均温度。

现在,至于这是否有好处……这取决于。它侧重于postgresql的plgpsql-execution部分而不是database-access部分,我自己的经验是plpgsql并不快。如果您可以轻松地对表进行查找以查找每个观察的前 15 分钟的行,那么自联接(如@danihp 答案)会做得很好。然而,这种方法可以处理来自一些更复杂的来源的观察,这些查找是不切实际的。与以往一样,在您自己的系统上试用和比较。

-- based on using this table definition
create table observation(id int primary key, timestamps timestamp not null unique,
                         temperature numeric(5,2) not null);

-- note that I'm reusing the table structure as a type for the state here
create type rollavg_state as (memory observation[], total numeric(5,2));

create function rollavg_func(state rollavg_state, next_in observation) returns rollavg_state immutable language plpgsql as $$
declare
  cutoff timestamp;
  i int;
  updated_memory observation[];
begin
  raise debug 'rollavg_func: state=%, next_in=%', state, next_in;
  cutoff := next_in.timestamps - '15 minutes'::interval;
  i := array_lower(state.memory, 1);
  raise debug 'cutoff is %', cutoff;
  while i <= array_upper(state.memory, 1) and state.memory[i].timestamps < cutoff loop
    raise debug 'shifting %', state.memory[i].timestamps;
    i := i + 1;
    state.total := state.total - state.memory[i].temperature;
  end loop;
  state.memory := array_append(state.memory[i:array_upper(state.memory, 1)], next_in);
  state.total := coalesce(state.total, 0) + next_in.temperature;
  return state;
end
$$;

create function rollavg_output(state rollavg_state) returns float8 immutable language plpgsql as $$
begin
  raise debug 'rollavg_output: state=% len=%', state, array_length(state.memory, 1);
  if array_length(state.memory, 1) > 0 then
    return state.total / array_length(state.memory, 1);
  else
    return null;
  end if;
end
$$;

create aggregate rollavg(observation) (sfunc = rollavg_func, finalfunc = rollavg_output, stype = rollavg_state);

-- referring to just a table name means a tuple value of the row as a whole, whose type is the table type
-- the aggregate relies on inputs arriving in ascending timestamp order
select rollavg(observation) over (order by timestamps) from observation;
于 2012-12-11T12:28:24.047 回答